leetcode单调栈数组-数组去重

import java.util.Stack;

/**
<p>给你一个字符串 <code>s</code> ，请你去除字符串中重复的字母，使得每个字母只出现一次。需保证 <strong>返回结果的字典序最小</strong>（要求不能打乱其他字符的相对位置）。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong><code>s = "bcabc"</code>
<strong>输出<code>：</code></strong><code>"abc"</code>
</pre>

<p><strong>示例 2：</strong></p>

<pre>
<strong>输入：</strong><code>s = "cbacdcbc"</code>
<strong>输出：</strong><code>"acdb"</code></pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
	<li><code>s</code> 由小写英文字母组成</li>
</ul>

<p>&nbsp;</p>

<p><strong>注意：</strong>该题与 1081 <a href="https://leetcode-cn.com/problems/smallest-subsequence-of-distinct-characters">https://leetcode-cn.com/problems/smallest-subsequence-of-distinct-characters</a> 相同</p>
<div><div>Related Topics</div><div><li>栈</li><li>贪心</li><li>字符串</li><li>单调栈</li></div></div><br><div><li>👍 715</li><li>👎 0</li></div>
*/

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public String removeDuplicateLetters(String s) {
        Stack<Character> stack = new Stack<>();

        //记录每个字符出现次数，后续排序会用到
        int[] count =new int[256];
        for(Character c :s.toCharArray()){
            count[c]++;
        }

        boolean[] inStack = new boolean[256];
        for(Character c :s.toCharArray()){

            count[c]--;
            if(inStack[c]==true){
                continue;
            }

            //为了条件三的排序
            //会把只出现一次 且排序靠后的给去掉
            while(!stack.isEmpty() && stack.peek()>c){

                if(count[stack.peek()]==0){
                    break;
                }

                inStack[stack.peek()]=false;
                stack.pop();
            }

            stack.push(c);
            inStack[c]= true;
        }
        StringBuffer sb = new StringBuffer();
        while(!stack.isEmpty()){
            sb.append(stack.pop());
        }
        return sb.reverse().toString();
    }
}
//leetcode submit region end(Prohibit modification and deletion)